MHT CET202523 Apr 2025Morning ShiftMathematicsProperties of TrianglesActual
With usual notations, in A B C , the lengths of two sides are 10 cm and 9 cm respectively. If angles A, B, C are in A.P. then perimeter of ABC is
Options
- A24+2 6 ~cm
- B24+ 6 ~cm
- C24-2 6 ~cm
- D24- 6 ~cm
Correct answer
B. 24+ 6 ~cm
Step-by-step solution
Given that angles A, B, C of ABC are in arithmetic progression, 2B = A + C . Since A + B + C = 180^ , substitution yields 3B = 180^ , so B = 60^ . Sides a = 10 cm and b = 9 cm are given. By the Cosine Rule applied to side b , b^2 = a^2 + c^2 - 2ac B . Substituting known values, 81 = 100 + c^2 - 20c 1 2 , simplifying to c^2 - 10c + 19 = 0 . Solving the quadratic, c = 5 6 , both satisfying triangle inequalities and yielding perimeters P = 24 6 cm. For an acute triangle, P = 24 + 6 cm.